Chapter 10
Quantity Relationships
in Chemical Reactions
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
Section 10.1
Conversion Factors from
a Chemical Equation
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
Goal 1
Given a chemical equation, or a reaction for
which the equation is known, and the number
of moles of one species in the reaction,
calculate the number of moles of any other
species.
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
The coefficients in a chemical equation give us
the conversion factors to get from the number
of particles of one substance, grouped into moles,
to the number of particles of another
substance in a chemical change.
A2 + 3 B2
2 AB3
1 mol A2
1 mol A2
3 mol B2
3 mol B2
2 mol AB3
2 mol AB3
3 mol B2
2 mol AB3
2 mol AB3
1 mol A2
1 mol A2
3 mol B2
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
Example:
How many moles of oxygen are needed
to completely react with 2.34 moles
of methane (CH4) in a combustion reaction?
Carbon dioxide and steam are the products.
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
CH4
+ 2 O2
CO2
+ 2 H 2O
PER relationship
from equation:
1 mol CH4 PER 2 mol O2
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
GIVEN: 2.34 mol CH4
PER/PATH: mol CH4
2.34 mol CH4 X
WANTED: mol O2
1 mol CH4/2 mol O2
2 mol O2
1 mol CH4
mol O2
= 4.68 mol O2
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
Section 10.2
Mass-Mass
Stoichiometry
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
Goal 2
Given a chemical equation, or a reaction for
which the equation can be written, and the
number of grams or moles of one species in
the reaction, find the number of grams or
moles of any other species.
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
Stoichiometry:
The quantitative relationships between the
substances involved in a chemical reaction, established
by the equation for the reaction
A stoichiometry problem asks,
“How much or how many?”
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
Prerequisite Skills for Stoichiometry
Write chemical formulas
Ch 6
Calculate molar masses
from chemical formulas
Sect 7.4
Use molar masses to change
mass to moles and moles to mass
Sect 7.5
Write and balance
chemical equations
Ch 8
Use the equation to change from moles of
Sect 10.1
one species to moles of another
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
Fundamental Stoichiometry Pattern:
Given
Macroscopic
Measurable
Quantity
Moles of
Given
Quantity
Moles of
Wanted
Quantity
Wanted
Macroscopic
Measurable
Quantity
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
Mass–Mass Stoichiometry Pattern:
Given
Mass
Moles of
Given
Quantity
Moles of
Wanted
Quantity
Molar Mass Equation
g PER mol
mol PER mol
Wanted
Mass
Molar Mass
g PER mol
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
How to Solve a Stoichiometry Problem:
The Stoichiometry Path
Step 1: Change the mass of the
given species to moles.
Step 2: Change the moles of the given species
to moles of the wanted species.
Step 3: Change the moles of the wanted
species to mass.
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
Example:
How many grams of carbon dioxide are produced
when 10.0 g of methane, CH4, is burned?
GIVEN: 10.0 g CH4
WANTED: g CO2
CH4
CO2
+ 2 O2
PER/PATH: g CH4
mol CH4
+ 2 H2O
16.04 g CH4/mol CH4
1 mol CO2/1 mol CH4
44.01 g CO2/mol CO2
mol CO2
g CO2
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
PER/PATH: g CH4
mol CH4
16.04 g CH4/mol CH4
1 mol CO2/1 mol CH4
44.01 g CO2/mol CO2
10.0 g CH4
g CO2
1 mol CH4
X
X
16.04 g CH4
1 mol CO2
1 mol CH4
X
44.01 g CO2
mol CO2
= 27.4 g CO2
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
Section 9.3
Percent Yield
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
Goal 3
Given two of the following, or information from
which two of the following may be
determined, calculate the third: theoretical
yield, actual yield, percent yield.
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
The actual yield of a chemical reaction
is usually less than the theoretical yield
predicted by a stoichiometry calculation because:
• reactants may be impure
• the reaction may not go to completion
• other reactions may occur
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
Percentage yield expresses the ratio of
actual yield to theoretical yield:
% yield =
actual yield
X 100
theoretical yield
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
Example:
Determine the percentage yield if 6.97 grams of
ammonia is produced from the reaction of 6.22
grams of nitrogen with excess hydrogen.
STEP 1: Calculate the theoretical yield
STEP 2: Use the given actual yield
and the calculated theoretical yield
to find the percentage yield
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
GIVEN: 6.22 g N2
N2
+ 3 H2
PER/PATH: g N2
mol N2
WANTED: g NH3 (theo)
2 NH3
28.02 g N2/mol N2
2 mol NH3/1 mol N2
17.03 g NH3/mol NH3
mol NH3
g NH3
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
PER/PATH: g N2
mol N2
28.02 g N2/mol N2
2 mol NH3/1 mol N2
17.03 g NH3/mol NH3
6.22 g N2 X
1 mol N2
28.02 g N2
17.03 g NH3
mol NH3
X
mol NH3
g NH3
2 mol NH3
1 mol N2
X
= 7.56 g NH3 (theo)
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
% yield =
actual yield
X 100
theoretical yield
= 6.97 g NH3 (act) X 100
7.56 g NH3 (theo)
= 92.2%
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.
Once a percentage yield has been
determined for a reaction, it can be used in
stoichiometry calculations
For example, a 92% yield means
92 g (act) PER 100 g (theo)
Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.